What is the tangent of f(x)=5x^2-x^3,(1,4) ?

The tangent of f(x)=5x^2-x^3,(1,4) is y=7x-3

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tangent of f(x)=5x^2-x^3,(1,4)

tangent of

f (x) = 5 x^{2} - x^{3}, (1, 4)

y = 7 x - 3

Solution steps

Find the slope of

f (x) = 5 x^{2} - x^{3} : \frac{df ( x )}{d x} = 10 x - 3 x^{2}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 7

Find the line with slope m=

7

and passing through

(1, 4) : y = 7 x - 3

y = 7 x - 3

f^{^{'}} (x) = tan (x)

\int \frac{4 x ^{2} - 19 x - 27}{( x + 1 ) ^{2} ( x - 3 )} d x

lo g_{4} (x)

f (x) = 3 x - 2

\int_{0}^{π} x^{2} sin (n x) d x

What is the tangent of f(x)=5x^2-x^3,(1,4) ?
The tangent of f(x)=5x^2-x^3,(1,4) is y=7x-3